Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $p = \dfrac{t^2 - 11t + 24}{3t^2 - 24t} \div \dfrac{t - 3}{-6t - 24} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{t^2 - 11t + 24}{3t^2 - 24t} \times \dfrac{-6t - 24}{t - 3} $ First factor the quadratic. $p = \dfrac{(t - 3)(t - 8)}{3t^2 - 24t} \times \dfrac{-6t - 24}{t - 3} $ Then factor out any other terms. $p = \dfrac{(t - 3)(t - 8)}{3t(t - 8)} \times \dfrac{-6(t + 4)}{t - 3} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ (t - 3)(t - 8) \times -6(t + 4) } { 3t(t - 8) \times (t - 3) } $ $p = \dfrac{ -6(t - 3)(t - 8)(t + 4)}{ 3t(t - 8)(t - 3)} $ Notice that $(t - 8)$ and $(t - 3)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ -6\cancel{(t - 3)}(t - 8)(t + 4)}{ 3t(t - 8)\cancel{(t - 3)}} $ We are dividing by $t - 3$ , so $t - 3 \neq 0$ Therefore, $t \neq 3$ $p = \dfrac{ -6\cancel{(t - 3)}\cancel{(t - 8)}(t + 4)}{ 3t\cancel{(t - 8)}\cancel{(t - 3)}} $ We are dividing by $t - 8$ , so $t - 8 \neq 0$ Therefore, $t \neq 8$ $p = \dfrac{-6(t + 4)}{3t} $ $p = \dfrac{-2(t + 4)}{t} ; \space t \neq 3 ; \space t \neq 8 $